Question: Multiply the following complex numbers: $({-4-2i}) \cdot ({1+3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4-2i}) \cdot ({1+3i}) = $ $ ({-4} \cdot {1}) + ({-4} \cdot {3}i) + ({-2}i \cdot {1}) + ({-2}i \cdot {3}i) $ Then simplify the terms: $ (-4) + (-12i) + (-2i) + (-6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (-12 - 2)i - 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (-12 - 2)i - (-6) $ The result is simplified: $ (-4 + 6) + (-14i) = 2-14i $